2023 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

Rewrite the equation of the curve as \(y = 3x^2 - 8x^{-1}\). Differentiate with respect to \(x\): \(\frac{dy}{dx} = 6x + 8x^{-2} = 6x + \frac{8}{x^2}\). Substitute \(x = 2\) into the derivative to find the gradient: \(\frac{dy}{dx} = 6(2) + \frac{8}{2^2} = 12 + 2 = 14\).

M1: for differentiating at least one term correctly to get \(6x\) or \(8x^{-2}\). M1: for a complete method of substituting \(x = 2\) into their differentiated expression. A1: for the correct gradient of 14.

Calculate the numerator: \(4 \times 10^5 \times 6 \times 10^{-3} = 24 \times 10^2\). Divide by the denominator: \(\frac{24 \times 10^2}{1.5 \times 10^{-2}} = 16 \times 10^4\). Convert to standard form: \(16 \times 10^4 = 1.6 \times 10^5\).

M1: for simplifying the numerator to \(24 \times 10^2\) or \(2400\), or for evaluating \(24 / 1.5 = 16\). M1: for a step showing division of powers of 10, resulting in \(16 \times 10^4\) or \(160000\). A1: for the final correct standard form \(1.6 \times 10^5\).

First, identify the elements in sets \(A\) and \(B\) within the universal set \(\mathcal{E}\): \(A = \{2, 3, 5, 7, 11\}\) and \(B = \{1, 2, 3, 4, 6, 12\}\). Find \(A'\), which contains all elements in \(\mathcal{E}\) that are not in \(A\): \(A' = \{1, 4, 6, 8, 9, 10, 12\}\). Now find the intersection of \(A'\) and \(B\), which contains the elements common to both sets: \(A' \cap B = \{1, 4, 6, 12\}\).

M1: for listing the correct elements of either set \(B\) or set \(A'\) (with at most one error). M1: for a clear process of finding the intersection of \(A'\) and \(B\). A1: for the correct list of members \(\{1, 4, 6, 12\}\) (order of elements does not matter, brackets are optional).

Multiply both sides by \((3 + x)\): \(y(3 + x) = 5x - 2\). Expand the bracket: \(3y + xy = 5x - 2\). Rearrange terms to get all \(x\) terms on one side: \(3y + 2 = 5x - xy\). Factorise \(x\) from the right-hand side: \(3y + 2 = x(5 - y)\). Divide both sides by \((5 - y)\): \(x = \frac{3y + 2}{5 - y}\).

M1: for multiplying by \((3 + x)\) and expanding correctly: \(3y + xy = 5x - 2\). M1: for isolating all terms in \(x\) on one side and factorising: \(x(5 - y) = 3y + 2\). A1: for \(x = \frac{3y + 2}{5 - y}\) (or equivalent algebraic expression).

Let \(\theta\) be the angle that the ladder makes with the ground. The side opposite to \(\theta\) is the height up the wall, \(4.5\text{ m}\). The side adjacent to \(\theta\) is the distance from the base, \(1.8\text{ m}\). Using the tangent ratio: \(\tan(\theta) = \frac{4.5}{1.8} = 2.5\). Calculate the angle: \(\theta = \tan^{-1}(2.5) \approx 68.19859...^\circ\). Rounding to 1 decimal place gives \(68.2^\circ\).

M1: for setting up a correct trigonometric ratio, e.g. \(\tan(\theta) = \frac{4.5}{1.8}\). M1: for \(\theta = \tan^{-1}(2.5)\) or \(68.198...\). A1: for \(68.2\) (accept \(68.2^\circ\)).

A reduction of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price. Let the original price be \(P\). Write the equation: \(0.85 \times P = 272\). Solve for \(P\): \(P = \frac{272}{0.85} = 320\). The original price was \(\$320\).

M1: for recognizing that \(85\%\) corresponds to \(272\), e.g., writing \(0.85 \times P = 272\) or \(\frac{272}{85} \times 100\). M1: for a complete method to find the original price: \(\frac{272}{0.85}\). A1: for \(320\) (accept \(\$320\)).

Find the differences between consecutive terms: First differences: \(7, 11, 15, 19\). Second differences: \(4, 4, 4\). Since the second differences are constant at \(4\), the coefficient of \(n^2\) is \(\frac{4}{2} = 2\). Subtract \(2n^2\) from each term in the original sequence: For \(n = 1\): \(3 - 2(1^2) = 1\). For \(n = 2\): \(10 - 2(2^2) = 2\). For \(n = 3\): \(21 - 2(3^2) = 3\). The sequence of differences is \(1, 2, 3, \dots\), which has the linear \(n\)-th term \(n\). Combine both parts: \(2n^2 + n\).

M1: for finding second difference of 4, leading to a term of \(2n^2\). M1: for subtracting \(2n^2\) from the terms of the sequence to find the linear part, or setting up simultaneous equations. A1: for \(2n^2 + n\) (or equivalent, e.g. \(n(2n + 1)\)).

1. Find the number of green marbles: The ratio of blue to green is \(3:5\). Since there are 12 blue marbles, \(3\) parts \(= 12 \implies 1\) part \(= 4\). Number of green marbles \(= 5 \times 4 = 20\). 2. Find the total number of blue and green marbles: \(12 + 20 = 32\). 3. Find the total number of marbles in the bag: The probability of choosing a red marble is \(0.2\), so the probability of choosing a blue or green marble is \(1 - 0.2 = 0.8\). Therefore, \(0.8 \times T = 32\). \(T = \frac{32}{0.8} = 40\).

M1: for finding the number of green marbles (20) or the total blue and green marbles (32). M1: for a complete method to find the total, e.g., \(\frac{32}{0.8}\). A1: for 40.

First, rewrite the equation of the curve as: \(y = 3x^2 - 16x^{-1}\). Differentiating with respect to \(x\) gives: \(\frac{dy}{dx} = 6x - 16(-1)x^{-2} = 6x + \frac{16}{x^2}\). Substitute \(x = 2\) into the derivative: \(\frac{dy}{dx} = 6(2) + \frac{16}{2^2} = 12 + \frac{16}{4} = 12 + 4 = 16\).

M1 for differentiating at least one term correctly: \(6x\) or \(16x^{-2}\). M1 for the fully correct derivative \(\frac{dy}{dx} = 6x + \frac{16}{x^2}\). A1 for a final answer of 16.

The total probability of all outcomes is 1. Therefore, the probability of choosing either a blue or a yellow counter is: \(1 - 0.35 = 0.65\). The ratio of blue to yellow counters is \(3 : 2\), which means the probability of choosing a blue counter is \(\frac{3}{3 + 2} = \frac{3}{5}\) of the combined probability of blue and yellow. Probability of blue = \(\frac{3}{5} \times 0.65 = 0.6 \times 0.65 = 0.39\).

M1 for finding the remaining probability: \(1 - 0.35 = 0.65\). M1 for multiplying by the correct fraction: \(\frac{3}{5} \times 0.65\) or equivalent. A1 for 0.39 (or equivalent fraction, e.g. \(\frac{39}{100}\)).

First, expand the numerator using the difference of two squares: \((3\sqrt{5} - 1)(3\sqrt{5} + 1) = (3\sqrt{5})^2 - 1^2 = (9 \times 5) - 1 = 45 - 1 = 44\). Next, simplify the denominator: \(\sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11}\). Substitute these back into the expression: \(\frac{44}{2\sqrt{11}} = \frac{22}{\sqrt{11}}\). Rationalise the denominator by multiplying the numerator and denominator by \(\sqrt{11}\): \(\frac{22\sqrt{11}}{11} = 2\sqrt{11}\). Therefore, \(k = 2\).

M1 for simplifying the numerator to 44. M1 for simplifying \(\sqrt{44}\) to \(2\sqrt{11}\) or for rationalising the denominator: e.g. \(\frac{44\sqrt{44}}{44} = \sqrt{44}\). A1 for \(k = 2\).

First, factor out the highest common factor, which is 3: \(12x^2 - 75y^2 = 3(4x^2 - 25y^2)\). Next, recognise that \(4x^2 - 25y^2\) is a difference of two squares: \(4x^2 - 25y^2 = (2x)^2 - (5y)^2 = (2x - 5y)(2x + 5y)\). Putting it all together, we get: \(3(2x - 5y)(2x + 5y)\).

M1 for factorising out 3: \(3(4x^2 - 25y^2)\). M1 for identifying and applying the difference of two squares to \((4x^2 - 25y^2)\) as \((2x - 5y)(2x + 5y)\). A1 for the fully factorised expression: \(3(2x - 5y)(2x + 5y)\) (factors can be in any order).

First, we find the gradient of the line with equation \(24x - y = 7\). Rewriting the equation in the form \(y = mx + c\) gives \(y = 24x - 7\), so the gradient of this line is \(24\). Since the tangents at \(A\) and \(B\) are parallel to this line, the gradient of the curve at these points is also \(24\). Next, we find the derivative \(\frac{dy}{dx}\) of the curve's equation: \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting the derivative equal to \(24\) gives: \(3x^2 - 12x + 9 = 24\) which simplifies to \(3x^2 - 12x - 15 = 0\). Dividing both sides by \(3\) yields: \(x^2 - 4x - 5 = 0\). Factoring the quadratic expression gives: \((x - 5)(x + 1) = 0\). This gives two possible values for the \(x\)-coordinate: \(x = 5\) and \(x = -1\). Now, we find the corresponding \(y\)-coordinates by substituting these \(x\)-values back into the original curve equation: For \(x = 5\): \(y = (5)^3 - 6(5)^2 + 9(5) + 5 = 125 - 150 + 45 + 5 = 25\). For \(x = -1\): \(y = (-1)^3 - 6(-1)^2 + 9(-1) + 5 = -1 - 6 - 9 + 5 = -11\). Thus, the coordinates of the two points are \((5, 25)\) and \((-1, -11)\).

- M1: For differentiating the curve equation to get \(\frac{dy}{dx} = 3x^2 - 12x + 9\) (at least two terms correct).
- M1: For setting their \(\frac{dy}{dx} = 24\).
- M1: For solving the quadratic equation to find the two \(x\)-values, e.g., \(x = 5\) and \(x = -1\).
- M1: For substituting at least one of their \(x\)-values back into the original equation for \(y\).
- A1: For both correct coordinates \((5, 25)\) and \((-1, -11)\).

Let \(N\) be the total number of counters in the bag. The number of red counters is \(0.2N\). Let the number of blue counters be \(b\) and green counters be \(g\). The total number of counters is given by \(r + b + g = N\), so \(0.2N + b + g = N\) which simplifies to \(b + g = 0.8N\). We are given that there are 6 more blue counters than green counters, so \(b = g + 6\). Substituting this into the previous equation: \((g + 6) + g = 0.8N \Rightarrow 2g + 6 = 0.8N \Rightarrow g = 0.4N - 3\). This means the number of blue counters is \(b = 0.4N - 3 + 6 = 0.4N + 3\). The probability of choosing two blue counters without replacement is: \(P(\text{Blue, Blue}) = \frac{b}{N} \times \frac{b-1}{N-1} = \frac{11}{38}\). Substituting the expression for \(b\) in terms of \(N\): \(\frac{(0.4N + 3)(0.4N + 2)}{N(N-1)} = \frac{11}{38}\). Expanding the numerator: \(\frac{0.16N^2 + 2N + 6}{N^2 - N} = \frac{11}{38}\). Cross-multiplying gives: \(38(0.16N^2 + 2N + 6) = 11(N^2 - N)\) which simplifies to: \(6.08N^2 + 76N + 228 = 11N^2 - 11N \Rightarrow 4.92N^2 - 87N - 228 = 0\). Multiplying the entire equation by \(25\) to eliminate the decimals gives: \(123N^2 - 2175N - 5700 = 0\). Dividing by \(3\) yields: \(41N^2 - 725N - 1900 = 0\). We can solve this quadratic equation using the quadratic formula: \(N = \frac{725 \pm \sqrt{(-725)^2 - 4(41)(-1900)}}{2(41)} = \frac{725 \pm \sqrt{525625 + 311600}}{82} = \frac{725 \pm \sqrt{837225}}{82} = \frac{725 \pm 915}{82}\). Since \(N\) must be positive, we choose the positive root: \(N = \frac{1640}{82} = 20\). Therefore, there are 20 counters in total in the bag.

- M1: For expressing the number of blue counters in terms of \(N\), e.g., \(b = 0.4N + 3\).
- M1: For setting up the probability equation: \(\frac{b(b-1)}{N(N-1)} = \frac{11}{38}\).
- M1: For substituting the expression of \(b\) in terms of \(N\) into the probability equation.
- M1: For simplifying to form a quadratic equation, e.g., \(41N^2 - 725N - 1900 = 0\) or \(N(N-1) = 380\).
- A1: For the correct answer 20.

Let the first term of the arithmetic series be \(a\) and the common difference be \(d\). The formula for the \(n\)-th term of an arithmetic series is \(u_n = a + (n-1)d\). For the 3rd term: \(u_3 = a + 2d = 13 \quad \text{--- (Equation 1)}\). The formula for the sum of the first \(n\) terms is \(S_n = \frac{n}{2}[2a + (n-1)d]\). For the sum of the first 8 terms: \(S_8 = \frac{8}{2}[2a + 7d] = 152 \Rightarrow 4[2a + 7d] = 152 \Rightarrow 2a + 7d = 38 \quad \text{--- (Equation 2)}\). To solve these simultaneous equations, we can multiply Equation 1 by \(2\): \(2a + 4d = 26 \quad \text{--- (Equation 3)}\). Subtracting Equation 3 from Equation 2 gives: \(3d = 12 \Rightarrow d = 4\). Substituting \(d = 4\) back into Equation 1: \(a + 2(4) = 13 \Rightarrow a = 5\). Now we can find the sum of the first 20 terms: \(S_{20} = \frac{20}{2}[2a + 19d] = 10[2(5) + 19(4)] = 10[10 + 76] = 10 \times 86 = 860\).

- M1: For writing a correct equation for the 3rd term, e.g., \(a + 2d = 13\).
- M1: For writing a correct equation for the sum of the first 8 terms, e.g., \(4(2a + 7d) = 152\) or \(2a + 7d = 38\).
- M1: For a complete method to solve the simultaneous equations to find the value of \(d\) or \(a\).
- M1: For finding both \(a = 5\) and \(d = 4\).
- A1: For the correct sum of 860.

Let \(M\) be the midpoint of the base edge \(AB\). Since the pyramid is a right pyramid, the triangle \(VAB\) is an isosceles triangle with \(VA = VB = 14\text{ cm}\). The line \(VM\) is perpendicular to the base edge \(AB\) and represents the slant height of the face \(VAB\). In the right-angled triangle \(VAM\), the hypotenuse is \(VA = 14\text{ cm}\) and the base is \(AM = \frac{1}{2}AB = 6\text{ cm}\). By Pythagoras' theorem: \(VM^2 = VA^2 - AM^2 = 14^2 - 6^2 = 196 - 36 = 160\), so \(VM = \sqrt{160}\text{ cm}\). Next, let \(O\) be the center of the rectangular base \(ABCD\). The line segment \(OM\) is parallel to the edge \(BC\) and its length is half the length of \(BC\): \(OM = \frac{1}{2}BC = 4.5\text{ cm}\). The angle between the face \(VAB\) and the base \(ABCD\) is the angle \(\angle VMO\). In the right-angled triangle \(VOM\), the adjacent side to the angle is \(OM = 4.5\text{ cm}\) and the hypotenuse is \(VM = \sqrt{160}\text{ cm}\). Using the cosine ratio: \(\cos(\angle VMO) = \frac{OM}{VM} = \frac{4.5}{\sqrt{160}}\). Calculating this gives: \(\cos(\angle VMO) \approx 0.355756\). Thus, \(\angle VMO = \arccos(0.355756) \approx 69.157^\circ\). Correct to 1 decimal place, the angle is \(69.2^\circ\).

- M1: For identifying the midpoint \(M\) of \(AB\) and stating \(AM = 6\) or \(OM = 4.5\).
- M1: For using Pythagoras' theorem to find the slant height \(VM\), e.g., \(VM = \sqrt{14^2 - 6^2} = \sqrt{160}\text{ cm}\).
- M1: For identifying the right-angled triangle \(VOM\) and the correct position of the angle \(\angle VMO\).
- M1: For using a correct trigonometric ratio, e.g., \(\cos(\angle VMO) = \frac{4.5}{\sqrt{160}}\).
- A1: For the correct angle in the range 69.1 to 69.2.

First, we express the vector \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Using the path from \(A\) to \(B\): \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). Since \(P\) divides \(AB\) in the ratio \(3:2\), we have: \(\overrightarrow{AP} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\). Therefore, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\). We are given that \(\overrightarrow{OX} = k\overrightarrow{OP}\), so: \(\overrightarrow{OX} = k\left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) = \frac{2}{5}k\mathbf{a} + \frac{3}{5}k\mathbf{b} \quad \text{--- (Equation 1)}\). Next, we express \(\overrightarrow{OX}\) via another path using the line segment \(AQ\). Since \(OQ : QB = 2 : 1\), \(\overrightarrow{OQ} = \frac{2}{3}\mathbf{b}\). The vector \(\overrightarrow{AQ}\) is: \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{2}{3}\mathbf{b}\). Since \(X\) lies on the line \(AQ\), we can write \(\overrightarrow{AX} = m\overrightarrow{AQ}\) for some scalar \(m\). Then, \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + m\left(-\mathbf{a} + \frac{2}{3}\mathbf{b}\right) = (1 - m)\mathbf{a} + \frac{2}{3}m\mathbf{b} \quad \text{--- (Equation 2)}\). Since the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, we can equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from Equation 1 and Equation 2: For \(\mathbf{a}\): \(\frac{2}{5}k = 1 - m\). For \(\mathbf{b}\): \(\frac{3}{5}k = \frac{2}{3}m \Rightarrow m = \frac{9}{10}k\). Substituting the expression for \(m\) into the first equation: \(\frac{2}{5}k = 1 - \frac{9}{10}k\). Multiplying the entire equation by \(10\): \(4k = 10 - 9k \Rightarrow 13k = 10 \Rightarrow k = \frac{10}{13}\).

- M1: For finding \(\overrightarrow{OP} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\).
- M1: For finding \(\overrightarrow{AQ} = -\mathbf{a} + \frac{2}{3}\mathbf{b}\).
- M1: For writing two expressions for \(\overrightarrow{OX}\), one in terms of \(k\) and one in terms of another scalar, e.g., \(m\).
- M1: For equating coefficients to form a system of equations, e.g., \(\frac{2}{5}k = 1 - m\) and \(\frac{3}{5}k = \frac{2}{3}m\).
- A1: For the correct value of \(k = \frac{10}{13}\) (or equivalent).

From the first equation, we can express \(y\) in terms of \(x\): \(y = 3 - 2x\). Substituting this expression for \(y\) into the second equation: \(x^2 - 3x(3 - 2x) + 2(3 - 2x)^2 = 12\). Now, we expand the terms: \(x^2 - (9x - 6x^2) + 2(9 - 12x + 4x^2) = 12\) which simplifies to: \(x^2 - 9x + 6x^2 + 18 - 24x + 8x^2 = 12\). Combining like terms: \(15x^2 - 33x + 18 = 12\). Subtracting \(12\) from both sides: \(15x^2 - 33x + 6 = 0\). Dividing the entire equation by \(3\) to simplify: \(5x^2 - 11x + 2 = 0\). This quadratic equation can be factored as: \((5x - 1)(x - 2) = 0\). This gives two values for \(x\): \(x = 2\) or \(x = \frac{1}{5} = 0.2\). Now we find the corresponding values of \(y\) using \(y = 3 - 2x\): For \(x = 2\): \(y = 3 - 2(2) = -1\). For \(x = 0.2\): \(y = 3 - 2(0.2) = 2.6\). Therefore, the solutions are \(x = 2, y = -1\) and \(x = 0.2, y = 2.6\).

- M1: For rearranging the linear equation to get an expression for \(y\) (or \(x\)), e.g., \(y = 3 - 2x\).
- M1: For substituting this expression into the quadratic equation.
- M1: For expanding and simplifying to obtain a standard form quadratic equation, e.g., \(5x^2 - 11x + 2 = 0\).
- M1: For solving the quadratic equation to find two values of \(x\) (or \(y\)).
- A1: For both pairs of solutions: \(x = 2, y = -1\) and \(x = 0.2, y = 2.6\) (or equivalent fractional forms).

Between 1st January 2022 and 1st January 2023, the investment grew from £10,609 to £10,927.27. This represents exactly 1 year of compound interest. Let the annual multiplier be \(M = 1 + \frac{r}{100}\). Therefore: \(10609 \times M = 10927.27 \Rightarrow M = \frac{10927.27}{10609} = 1.03\). Since \(M = 1.03\), the interest rate is: \(r = (1.03 - 1) \times 100 = 3\). Next, we find the initial investment \(P\). From 1st January 2020 to 1st January 2022 is a period of 2 years. Using the compound interest formula: \(P \times (1.03)^2 = 10609\). Since \(1.03^2 = 1.0609\), we have: \(P \times 1.0609 = 10609 \Rightarrow P = \frac{10609}{1.0609} = 10000\). Thus, the values are \(r = 3\) and \(P = 10000\).

- M1: For setting up an equation to find the multiplier for one year, e.g., \(10609 \times (1 + r/100) = 10927.27\).
- A1: For \(r = 3\) (or multiplier \(1.03\)).
- M1: For setting up the 2-year growth equation for \(P\), e.g., \(P \times 1.03^2 = 10609\).
- M1: For solving the equation to find \(P\), e.g., \(P = \frac{10609}{1.0609}\).
- A1: For \(P = 10000\).

First, we know that \(A, B, C, D\) are points on a circle, so the quadrilateral \(ABCD\) is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to \(180^\circ\). Therefore, we can find the angle \(\angle ABC\) in terms of \(x\): \(\angle ABC = 180^\circ - \angle ADC = 180^\circ - x^\circ\). Next, the angle subtended by an arc at the center of a circle is twice the angle subtended at the circumference. The obtuse angle \(AOC\) is the angle at the center subtended by the arc \(ADC\), and \(\angle ABC\) is the angle at the circumference. Therefore: \(\text{obtuse angle } AOC = 2 \times \angle ABC = 2(180 - x)^\circ = (360 - 2x)^\circ\). We are given that the obtuse angle \(AOC = (3x - 110)^\circ\). Equating the two expressions: \(3x - 110 = 360 - 2x\). Adding \(2x\) to both sides: \(5x - 110 = 360\). Adding \(110\) to both sides: \(5x = 470\). Dividing by \(5\): \(x = 94\).

- M1: For expressing angle \(ABC\) as \(180 - x\) or the reflex angle \(AOC\) as \(2x\).
- M1: For establishing the link between the obtuse angle and circumference, e.g., \(AOC = 2(180 - x)\) or \(AOC = 360 - 2x\).
- M1: For setting up the equation \(3x - 110 = 360 - 2x\).
- A1: For \(x = 94\).
- C1: For providing both reasons: 'opposite angles of a cyclic quadrilateral sum to 180 degrees' AND 'the angle at the center is twice the angle at the circumference'.

Since \(P(1, -10)\) lies on the curve, substitute \(x = 1\) and \(y = -10\): \(-10 = a(1)^3 + b(1)^2 - 24(1) + 11 \implies a + b = 3\) (Equation 1). Next, find the derivative \(\frac{dy}{dx} = 3ax^2 + 2bx - 24\). Since \(P\) is a stationary point, \(\frac{dy}{dx} = 0\) when \(x = 1\): \(3a(1)^2 + 2b(1) - 24 = 0 \implies 3a + 2b = 24\) (Equation 2). Solve the simultaneous equations: Multiply Equation 1 by 2: \(2a + 2b = 6\). Subtract this from Equation 2: \((3a + 2b) - (2a + 2b) = 24 - 6 \implies a = 18\). Substitute \(a = 18\) into Equation 1: \(18 + b = 3 \implies b = -15\). To find the other stationary point, substitute \(a = 18\) and \(b = -15\) into \(\frac{dy}{dx} = 0\): \(54x^2 - 30x - 24 = 0\). Divide by 6: \(9x^2 - 5x - 4 = 0\). Factorise: \((9x + 4)(x - 1) = 0\). Therefore, \(x = 1\) or \(x = -\frac{4}{9}\). The \(x\)-coordinate of the other stationary point is \(-\frac{4}{9}\).

M1: Substituting \(x=1, y=-10\) into the curve equation to get \(a + b = 3\) (or equivalent). M1: Differentiating the curve equation to obtain \(\frac{dy}{dx} = 3ax^2 + 2bx - 24\). M1: Setting \(\frac{dy}{dx} = 0\) at \(x = 1\) to get \(3a + 2b = 24\), and solving simultaneously to find \(a = 18\) and \(b = -15\). M1: Substituting \(a\) and \(b\) back into \(\frac{dy}{dx} = 0\) to form the quadratic equation \(54x^2 - 30x - 24 = 0\) (or equivalent). A1: Finding \(x = -\frac{4}{9}\) (or equivalent, e.g., \(-0.\dot{4}\)).

First, calculate the length of the diagonal \(AC\) of the rectangular base using Pythagoras' Theorem: \(AC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13\text{ cm}\). Since \(O\) is the center of the base, the distance \(AO = \frac{13}{2} = 6.5\text{ cm}\). In the right-angled triangle \(VOA\), the height \(VO\) of the pyramid is given by: \(VO = 6.5 \times \tan(62^\circ) \approx 12.2247\text{ cm}\). Let \(M\) be the midpoint of \(BC\). The line \(OM\) is perpendicular to \(BC\) and its length is \(OM = \frac{1}{2} AB = 6\text{ cm}\). The angle between the face \(VBC\) and the base is the angle \(\angle VMO\) in the right-angled triangle \(VOM\): \(\tan(\angle VMO) = \frac{VO}{OM} = \frac{12.2247}{6} \approx 2.03745\). Thus, \(\angle VMO = \arctan(2.03745) \approx 63.856^\circ\). Rounded to 1 decimal place, the angle is \(63.9^\circ\).

M1: Finding the diagonal of the base \(AC = 13\text{ cm}\) or half-diagonal \(AO = 6.5\text{ cm}\). M1: Correct trigonometric expression to find the vertical height: \(VO = 6.5 \tan(62^\circ)\) (approx. \(12.22\text{ cm}\)). M1: Identifying the required angle as \(\angle VMO\), where \(M\) is the midpoint of \(BC\). M1: Stating \(OM = 6\text{ cm}\) and using \(\tan(\angle VMO) = \frac{VO}{6}\). A1: Correct final angle of \(63.9^\circ\) (accept \(63.8^\circ\) to \(64.0^\circ\)).

The probability of choosing two red counters without replacement is: \(P(\text{Red, Red}) = \frac{9}{N} \times \frac{8}{N-1} = \frac{72}{N(N-1)}\). We are given this probability is \(\frac{6}{13}\), so: \(\frac{72}{N(N-1)} = \frac{6}{13}\). Rearranging gives: \(72 \times 13 = 6N(N-1) \implies 936 = 6(N^2 - N)\). Dividing by 6 yields: \(N^2 - N - 156 = 0\). Factorising the quadratic equation: \((N - 13)(N + 12) = 0\). Since the number of counters must be positive, \(N = 13\). The number of red counters is 9, so the number of blue counters is \(13 - 9 = 4\).

M1: Writing an algebraic expression for selecting two red counters: \(\frac{9}{N} \times \frac{8}{N-1}\). M1: Creating the equation \(\frac{72}{N(N-1)} = \frac{6}{13}\). M1: Rearranging to form a standard quadratic equation: \(N^2 - N - 156 = 0\) (or equivalent). M1: Solving the quadratic equation to find \(N = 13\) (rejecting \(N = -12\)). A1: Deducing the correct number of blue counters, \(4\).

(a) To find \(f^{-1}(x)\), let \(y = \frac{3x + 2}{x - 1}\). Multiply both sides by \(x - 1\): \(y(x - 1) = 3x + 2 \implies yx - y = 3x + 2\). Group all \(x\) terms on one side: \(yx - 3x = y + 2 \implies x(y - 3) = y + 2\). Make \(x\) the subject: \(x = \frac{y + 2}{y - 3}\). Therefore, \(f^{-1}(x) = \frac{x + 2}{x - 3}\). (b) Since \(fg(x) = 8\), we can write \(g(x) = f^{-1}(8)\). Using the expression from part (a): \(f^{-1}(8) = \frac{8 + 2}{8 - 3} = \frac{10}{5} = 2\). Thus, \(2x - 5 = 2 \implies 2x = 7 \implies x = 3.5\).

M1: (Part a) Writing \(y = \frac{3x+2}{x-1}\) and multiplying by \(x-1\) to get \(y(x-1) = 3x+2\). M1: (Part a) Grouping \(x\) terms and factorising: \(x(y-3) = y+2\). A1: (Part a) Correct expression \(f^{-1}(x) = \frac{x+2}{x-3}\). M1: (Part b) Finding \(f^{-1}(8)\) or substituting \(g(x)\) into \(f(x)\) to get \(\frac{3(2x-5)+2}{2x-6} = 8\). A1: (Part b) Finding \(x = 3.5\) (or \(\frac{7}{2}\)).

Multiply the entire equation by the common denominator \((x - 1)(x + 1)\) to clear the fractions: \(6(x + 1) - 3(x - 1) = 1(x - 1)(x + 1)\). Expand both sides: \(6x + 6 - 3x + 3 = x^2 - 1 \implies 3x + 9 = x^2 - 1\). Rearrange the equation into a standard quadratic form: \(x^2 - 3x - 10 = 0\). Factorise the quadratic: \((x - 5)(x + 2) = 0\). This gives the solutions: \(x = 5\) or \(x = -2\).

M1: For multiplying by the common denominator to obtain \(6(x+1) - 3(x-1) = (x-1)(x+1)\) (or equivalent). M1: Expanding correctly to get \(3x + 9\) on one side and \(x^2 - 1\) on the other. M1: Rearranging terms to form the standard quadratic equation \(x^2 - 3x - 10 = 0\). M1: Factorising as \((x-5)(x+2) = 0\) (or using the quadratic formula with at most one sign error). A1: Both correct values \(x = 5\) and \(x = -2\).

First, write the equation of the curve in a form that is easy to differentiate:
\(y = 4x^3 - 6x^{-2} + 5\)

Now, differentiate with respect to \(x\):
\(\frac{dy}{dx} = 12x^2 - 6(-2)x^{-3}\)
\(\frac{dy}{dx} = 12x^2 + 12x^{-3}\)
\(\frac{dy}{dx} = 12x^2 + \frac{12}{x^3}\)

Substitute \(x = 2\) into the derivative to find the gradient:
\(\text{Gradient} = 12(2)^2 + \frac{12}{2^3}\)
\(\text{Gradient} = 12(4) + \frac{12}{8}\)
\(\text{Gradient} = 48 + 1.5 = 49.5\)

M1: For attempt to differentiate, with at least one term correct (e.g., \(12x^2\) or \(12x^{-3}\)).
M1: For the complete correct derivative \(\frac{dy}{dx} = 12x^2 + 12x^{-3}\).
A1: For \(49.5\) (or equivalent fraction, e.g., \(\frac{99}{2}\) or \(49\frac{1}{2}\)).

First, find an expression for the composite function \(fg(x)\) by substituting \(g(x)\) into \(f(x)\):
\(fg(x) = f(2x + 1) = \frac{4}{(2x + 1) - 3}\)
\(fg(x) = \frac{4}{2x - 2}\)

Set this expression equal to \(\frac{1}{2}\):
\(\frac{4}{2x - 2} = \frac{1}{2}\)

Solve for \(x\) by cross-multiplying:
\(4 \times 2 = 1 \times (2x - 2)\)
\(8 = 2x - 2\)
\(10 = 2x\)
\(x = 5\)

M1: For a correct expression for \(fg(x)\), e.g., \(\frac{4}{2x+1-3}\) or \(\frac{4}{2x-2}\).
M1: For setting their \(fg(x) = \frac{1}{2}\) and attempting to solve, leading to a linear equation such as \(8 = 2x - 2\).
A1: For \(x = 5\) (with working shown).

The total number of counters in the bag is \(5 + 7 = 12\).
We want to find the probability of choosing two counters of different colours. This can happen in two ways:
1. Red then Blue (RB)
2. Blue then Red (BR)

Calculate the probability of Red then Blue:
\(P(RB) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)

Calculate the probability of Blue then Red:
\(P(BR) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\)

Add the two probabilities together to find the total probability of different colours:
\(P(\text{different colours}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132}\)

Simplify the fraction:
\(\frac{70}{132} = \frac{35}{66}\)

M1: For calculating the probability of one correct combination, e.g., \(\frac{5}{12} \times \frac{7}{11}\) or \(\frac{7}{12} \times \frac{5}{11}\).
M1: For adding both correct probabilities: \(\frac{5}{12} \times \frac{7}{11} + \frac{7}{12} \times \frac{5}{11}\).
A1: For \(\frac{35}{66}\) (or equivalent fraction, e.g., \(\frac{70}{132}\); accept decimal \(0.530\) or better).

Express both 27 and 9 as powers of 3:
\(27 = 3^3\)
\(9 = 3^2\), which means \(\frac{1}{9^{x+2}} = 9^{-(x+2)} = (3^2)^{-(x+2)} = 3^{-2(x+2)}\)

Substitute these back into the equation:
\((3^3)^{2x-1} = 3^{-2(x+2)}\)
\(3^{3(2x-1)} = 3^{-2(x+2)}\)
\(3^{6x-3} = 3^{-2x-4}\)

Since the bases are now the same, we can equate the indices:
\(6x - 3 = -2x - 4\)

Solve for \(x\):
\(8x - 3 = -4\)
\(8x = -1\)
\(x = -\frac{1}{8}\)

M1: For expressing 27 as \(3^3\) or 9 as \(3^2\) (or \(1/9\) as \(3^{-2}\)).
M1: For equating indices to form a correct linear equation, e.g., \(3(2x-1) = -2(x+2)\) or \(6x - 3 = -2x - 4\).
A1: For \(-\frac{1}{8}\) (or \(-0.125\)).

Let's find the first differences between successive terms:
\(10 - 3 = 7\)
\(21 - 10 = 11\)
\(36 - 21 = 15\)
\(55 - 36 = 19\)
First differences: \(7, 11, 15, 19\)

Now, find the second differences:
\(11 - 7 = 4\)
\(15 - 11 = 4\)
\(19 - 15 = 4\)
Second differences: \(4, 4, 4\)

Since the second difference is constant at 4, the coefficient of the \(n^2\) term is \(\frac{4}{2} = 2\).
Thus, the quadratic part of the sequence is \(2n^2\).

Subtract \(2n^2\) from each term in the original sequence to find the linear remainder:
- For \(n=1\): \(3 - 2(1)^2 = 3 - 2 = 1\)
- For \(n=2\): \(10 - 2(2)^2 = 10 - 8 = 2\)
- For \(n=3\): \(21 - 2(3)^2 = 21 - 18 = 3\)
- For \(n=4\): \(36 - 2(4)^2 = 36 - 32 = 4\)
- For \(n=5\): \(55 - 2(5)^2 = 55 - 50 = 5\)

The remaining sequence is \(1, 2, 3, 4, 5, \dots\), which has the \(n\)-th term of \(n\).

Combining these parts, the \(n\)-th term of the sequence is \(2n^2 + n\).

M1: For finding the constant second difference is 4, or identifying the term \(2n^2\).
M1: For a full and correct method to find the linear component (e.g., subtracting \(2n^2\) from sequence terms to obtain \(1, 2, 3, \dots\), or setting up simultaneous equations).
A1: For \(2n^2 + n\) (or equivalent, e.g., \(n(2n+1)\)).

In the right-angled triangle \(ABC\):
\(\tan(28^\circ) = \frac{AB}{BC} = \frac{24}{BC}\)
\(BC = \frac{24}{\tan(28^\circ)} \approx 45.136\text{ m}\)

In the right-angled triangle \(ABD\):
\(\tan(43^\circ) = \frac{AB}{BD} = \frac{24}{BD}\)
\(BD = \frac{24}{\tan(43^\circ)} \approx 25.737\text{ m}\)

Since \(C\), \(D\), and \(B\) lie on a straight line, and \(C\) and \(D\) are on the same side of \(B\):
\(CD = BC - BD\)
\(CD = 45.136 - 25.737 = 19.399\text{ m}\)

Rounding to 3 significant figures gives \(19.4\text{ m}\).

M1: For a correct trigonometric equation to find either \(BC\) or \(BD\), e.g., \(\tan(28^\circ) = \frac{24}{BC}\) or \(\tan(43^\circ) = \frac{24}{BD}\).
M1: For finding both \(BC \approx 45.1\) to \(45.2\) and \(BD \approx 25.7\) to \(25.8\).
A1: For \(19.4\) (accept answers in the range \(19.3\) to \(19.5\)).

The volume of a sphere of radius \(r\) is:
\(V_{\text{sphere}} = \frac{4}{3} \pi r^3\)

The volume of a cone of base radius \(r\) and height \(h\) is:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\)

Since no metal is lost, their volumes are equal:
\(\frac{4}{3} \pi r^3 = \frac{1}{3} \pi r^2 h\)

We can simplify this equation by dividing both sides by \(\frac{1}{3} \pi r^2\) (since \(r \neq 0\)):
\(4r = h\)

To write this as the ratio \(r : h\):
\(r : h = r : 4r = 1 : 4\)

Thus, \(n = 4\) and the ratio is \(1 : 4\).

M1: For setting up an equation equating the correct volume formulas, e.g., \(\frac{4}{3} \pi r^3 = \frac{1}{3} \pi r^2 h\).
M1: For simplifying the equation to find a linear relation between \(r\) and \(h\), e.g., \(4r = h\) or \(\frac{r}{h} = \frac{1}{4}\).
A1: For \(1:4\) (accept \(r:h = 1:4\)).

Multiply both sides by \(5 - y\) to clear the fraction:
\(w(5 - y) = 3 + 2y\)

Expand the brackets:
\(5w - wy = 3 + 2y\)

Rearrange the equation to group all terms containing \(y\) on one side, and all other terms on the opposite side:
\(5w - 3 = 2y + wy\)

Factorise \(y\) from the right-hand side:
\(5w - 3 = y(2 + w)\)

Divide both sides by \(2 + w\) to isolate \(y\):
\(y = \frac{5w - 3}{2 + w}\)

M1: For clearing the fraction by multiplying by \((5-y)\), i.e., \(w(5-y) = 3+2y\).
M1: For expanding and gathering terms in \(y\) on one side and terms without \(y\) on the other, e.g., \(5w-3 = wy + 2y\) or \(3-5w = -2y-wy\).
A1: For \(y = \frac{5w - 3}{w + 2}\) (or equivalent, e.g., \(y = \frac{3 - 5w}{-w - 2}\)).

To find the turning points, we first find the first derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x^2 - 18x - 24\). At a turning point, \(\frac{dy}{dx} = 0\): \(6x^2 - 18x - 24 = 0\). Divide the equation by 6: \(x^2 - 3x - 4 = 0\). Factorise the quadratic equation: \((x - 4)(x + 1) = 0\). This gives \(x = 4\) or \(x = -1\). Since we are given that \(x > 0\), we must have \(x = 4\). Now, substitute \(x = 4\) back into the original curve equation to find the \(y\)-coordinate: \(y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105\). Therefore, the coordinates of the turning point are \((4, -105)\).

M1: for differentiating to get \(6x^2 - 18x - 24\) (at least two terms correct). M1: for setting their \(\frac{dy}{dx} = 0\) and solving to find \(x = 4\) (ignoring \(x = -1\)). A1: for \((4, -105)\) or \(x = 4, y = -105\).

The total number of counters in the bag is \(8 + n\). The probability of choosing a red counter first is \(\frac{8}{8+n}\). Since the first counter is not replaced, there are now 7 red counters remaining, and the total number of counters is \(7 + n\). The probability of choosing a red counter second is \(\frac{7}{7+n}\). The probability that both counters are red is: \(\frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{39}\). This simplifies to: \(\frac{56}{(8+n)(7+n)} = \frac{14}{39}\). Dividing both sides by 14 gives: \(\frac{4}{(8+n)(7+n)} = \frac{1}{39}\). Multiplying both sides by \(39(8+n)(7+n)\) gives: \(156 = (8+n)(7+n)\) which expands to \(156 = 56 + 15n + n^2\). Rearranging gives: \(n^2 + 15n - 100 = 0\). Factorising this quadratic equation: \((n - 5)(n + 20) = 0\). Since \(n\) must be a positive integer, we have \(n = 5\).

M1: for setting up the probability equation \(\frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{39}\). M1: for expanding and simplifying to a 3-term quadratic equation, e.g., \(n^2 + 15n - 100 = 0\). A1: for \(5\) (rejecting \(-20\)).

We can rewrite the first term of the equation as \(3 \cdot (3^x)^2\). Let \(y = 3^x\). The equation becomes: \(3y^2 - 10y + 3 = 0\). Factorising this quadratic equation: \((3y - 1)(y - 3) = 0\), which gives \(y = \frac{1}{3}\) or \(y = 3\). Now, substitute \(y = 3^x\) back into these solutions. Case 1: If \(3^x = \frac{1}{3}\), then \(3^x = 3^{-1}\), which gives \(x = -1\). Case 2: If \(3^x = 3\), then \(3^x = 3^1\), which gives \(x = 1\). Thus, the solutions are \(x = -1\) and \(x = 1\).

M1: for substituting \(y = 3^x\) to obtain a quadratic equation \(3y^2 - 10y + 3 = 0\) (or equivalent). M1: for finding the solutions \(y = \frac{1}{3}\) and \(y = 3\). A1: for \(x = -1\) and \(x = 1\) (both required).

Let \(y = f(x)\), so \(y = \frac{3x+2}{x-5}\). Multiply both sides by \(x-5\) to get: \(y(x-5) = 3x+2\). Expanding the left side gives: \(xy - 5y = 3x+2\). Rearrange the equation to group all terms with \(x\) on one side: \(xy - 3x = 5y + 2\). Factorise out \(x\) on the left side: \(x(y - 3) = 5y + 2\). Divide by \(y - 3\) to isolate \(x\): \(x = \frac{5y+2}{y-3}\). Replacing \(y\) with \(x\), we get the inverse function: \(f^{-1}(x) = \frac{5x+2}{x-3}\).

M1: for starting with \(y = \frac{3x+2}{x-5}\) and multiplying both sides by \(x-5\). M1: for expanding and factorising to isolate \(x\), e.g., \(x(y-3) = 5y+2\). A1: for \(\frac{5x+2}{x-3}\) (or equivalent form).

We use the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} ac \sin B\). Substituting the given values: \(21\sqrt{3} = \frac{1}{2} \times 12 \times 7 \times \sin B\), which simplifies to \(21\sqrt{3} = 42 \sin B\). Solving for \(\sin B\) gives: \(\sin B = \frac{21\sqrt{3}}{42} = \frac{\sqrt{3}}{2}\). Since angle \(ABC\) is obtuse (between \(90^\circ\) and \(180^\circ\)), we have \(B = 120^\circ\). Now, we use the cosine rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B\). Substituting the values: \(AC^2 = 7^2 + 12^2 - 2(7)(12)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we get: \(AC^2 = 49 + 144 - 168(-0.5) = 193 + 84 = 277\). Thus, the exact length of \(AC\) is \(\sqrt{277}\text{ cm}\).

M1: for setting up the area equation \(\frac{1}{2} \times 12 \times 7 \times \sin B = 21\sqrt{3}\) and finding \(B = 120^\circ\) (or \(\sin B = \frac{\sqrt{3}}{2}\) with mention of obtuse angle). M1: for substituting their angle and side lengths into the cosine rule formula \(AC^2 = 7^2 + 12^2 - 2 \times 7 \times 12 \times \cos(120^\circ)\). A1: for \(\sqrt{277}\).

(a) The total number of counters in the bag is \(r + 4\).

The probability of selecting a red counter first is \(\frac{r}{r+4}\).

After removing one red counter, there are \(r - 1\) red counters left and the total number of counters becomes \(r + 3\).

The probability of selecting a second red counter is \(\frac{r-1}{r+3}\).

Since the probability of both counters being red is \(\frac{1}{3}\):
\(\frac{r}{r+4} \times \frac{r-1}{r+3} = \frac{1}{3}\)

\(\frac{r(r-1)}{(r+4)(r+3)} = \frac{1}{3}\)

Multiply both sides by \(3(r+4)(r+3)\):
\(3r(r-1) = (r+4)(r+3)\)

\(3r^2 - 3r = r^2 + 7r + 12\)

Rearranging all terms to one side:
\(3r^2 - r^2 - 3r - 7r - 12 = 0\)

\(2r^2 - 10r - 12 = 0\)

(b) We can simplify the quadratic equation by dividing by 2:
\(r^2 - 5r - 6 = 0\)

Factorising the quadratic equation:
\((r - 6)(r + 1) = 0\)

This gives \(r = 6\) or \(r = -1\).

Since the number of counters must be positive, \(r = 6\).

(a)
- M1: For writing an expression for the probability of selecting two red counters: \(\frac{r}{r+4} \times \frac{r-1}{r+3}\).
- M1: For setting this equal to \(\frac{1}{3}\) and multiplying to clear fractions: \(3r(r-1) = (r+4)(r+3)\).
- A1: For correctly expanding both sides and showing \(2r^2 - 10r - 12 = 0\) with no algebraic errors.

(b)
- M1: For a method to solve the quadratic equation, e.g., factorising to \((r-6)(r+1) = 0\) or using the quadratic formula.
- A1: For \(r = 6\) as the only valid solution (must reject \(r = -1\)).

(a) The velocity \(v\) of the particle is the rate of change of displacement with respect to time:
\(v = \frac{\text{d}s}{\text{d}t} = 3t^2 - 12t - 15\)

The particle is momentarily at rest when \(v = 0\):
\(3t^2 - 12t - 15 = 0\)

Dividing the entire equation by 3:
\(t^2 - 4t - 5 = 0\)

Factorising the quadratic expression:
\((t - 5)(t + 1) = 0\)

This gives \(t = 5\) or \(t = -1\).

Since time \(t \ge 0\), the particle is at rest when \(t = 5\) seconds.

(b) The acceleration \(a\) is the rate of change of velocity with respect to time:
\(a = \frac{\text{d}v}{\text{d}t} = 6t - 12\)

Substituting \(t = 5\) into the acceleration expression:
\(a = 6(5) - 12 = 30 - 12 = 18\text{ m/s}^2\).

(a)
- M1: For differentiating \(s\) to get an expression for velocity: \(3t^2 - 12t - 15\).
- M1: For setting their velocity expression equal to 0 and attempting to solve the quadratic equation.
- A1: For \(t = 5\) (and omitting or rejecting \(t = -1\)).

(b)
- M1: For differentiating their velocity expression to find acceleration: \(6t - 12\).
- A1: For substituting \(t = 5\) and getting a final answer of \(18\).

(a) The volume of a sphere is given by \(V = \frac{4}{3}\pi R^3\).
For a radius of \(3r\):
\(V_{\text{sphere}} = \frac{4}{3}\pi (3r)^3 = \frac{4}{3}\pi (27r^3) = 36\pi r^3\)

The volume of a cone is given by \(V = \frac{1}{3}\pi R^2 h\).
For a base radius of \(2r\):
\(V_{\text{cone}} = \frac{1}{3}\pi (2r)^2 h = \frac{4}{3}\pi r^2 h\)

We are given that \(V_{\text{sphere}} = 3 \times V_{\text{cone}}\):
\(36\pi r^3 = 3 \left(\frac{4}{3}\pi r^2 h\right)\)
\(36\pi r^3 = 4\pi r^2 h\)

Dividing both sides by \(4\pi r^2\):
\(h = 9r\)

(b) To find the total surface area of the cone, we first need its slant height \(l\).
Using Pythagoras' theorem:
\(l = \sqrt{(2r)^2 + h^2} = \sqrt{4r^2 + (9r)^2} = \sqrt{4r^2 + 81r^2} = \sqrt{85r^2} = r\sqrt{85}\)

The total surface area of the cone is the sum of the base area and the curved surface area:
\(A = \pi R^2 + \pi R l\)
\(A = \pi (2r)^2 + \pi (2r)(r\sqrt{85})\)
\(A = 4\pi r^2 + 2\sqrt{85}\pi r^2 = (4 + 2\sqrt{85})\pi r^2\)

Thus, \(a = 4\), \(b = 2\), and \(c = 85\).

(a)
- M1: For expressing the volume of the sphere as \(36\pi r^3\) or the volume of the cone as \(\frac{4}{3}\pi r^2 h\).
- A1: For setting up the equation \(36\pi r^3 = 4\pi r^2 h\) and showing the steps clearly to derive \(h = 9r\).

(b)
- M1: For a method to find the slant height \(l\) in terms of \(r\) using Pythagoras' theorem: \(l = \sqrt{(2r)^2 + (9r)^2}\).
- M1: For substituting \(R = 2r\) and their slant height into the formula \(\pi R^2 + \pi R l\).
- A1: For the exact final answer \((4 + 2\sqrt{85})\pi r^2\).

(a) To find the inverse function \(f^{-1}(x)\), let \(y = f(x)\):
\(y = \frac{3x + 1}{x - 2}\)

Multiply both sides by \((x - 2)\):
\(y(x - 2) = 3x + 1\)
\(yx - 2y = 3x + 1\)

Rearrange the equation to group all terms with \(x\) on one side:
\(yx - 3x = 2y + 1\)

Factorise \(x\):
\(x(y - 3) = 2y + 1\)

Divide by \(y - 3\):
\(x = \frac{2y + 1}{y - 3}\)

Replacing \(y\) with \(x\), we get:
\(f^{-1}(x) = \frac{2x + 1}{x - 3}\)

(b) To solve \(fg(x) = 4\), we can apply \(f^{-1}\) to both sides:
\(g(x) = f^{-1}(4)\)

Substitute \(x = 4\) into our inverse function:
\(f^{-1}(4) = \frac{2(4) + 1}{4 - 3} = \frac{8 + 1}{1} = 9\)

Now, set \(g(x) = 9\):
\(2x - 5 = 9\)
\(2x = 14\)
\(x = 7\)

(a)
- M1: For a correct start to find the inverse, such as writing \(y = \frac{3x+1}{x-2}\) and expanding to \(y(x-2) = 3x+1\).
- M1: For isolating \(x\) to get \(x(y-3) = 2y+1\) (or equivalent with swapped variables).
- A1: For the correct expression \(f^{-1}(x) = \frac{2x+1}{x-3}\).

(b)
- M1: For a correct method to solve the composite equation, such as finding \(fg(x) = \frac{3(2x-5)+1}{2x-5-2}\) and setting equal to 4, or setting \(2x-5 = f^{-1}(4)\).
- A1: For the final answer \(x = 7\).

(a) First, find the distance from corner \(A\) to the centre of the base \(O\).
The diagonal of the square base \(ABCD\) is:
\(AC = \sqrt{8^2 + 8^2} = \sqrt{128} = 8\sqrt{2}\text{ cm}\)

The distance \(AO\) is half of the diagonal:
\(AO = \frac{1}{2} AC = 4\sqrt{2}\text{ cm} \approx 5.657\text{ cm}\)

In the right-angled triangle \(VOA\):
\(VA = \sqrt{VO^2 + AO^2} = \sqrt{12^2 + (4\sqrt{2})^2} = \sqrt{144 + 32} = \sqrt{176} \approx 13.266\text{ cm}\)

To 3 significant figures, \(VA = 13.3\text{ cm}\).

(b) Let \(M\) be the midpoint of the base side \(AB\).
Since \(O\) is the centre of the square base of side length 8 cm, the distance from \(O\) to the midpoint \(M\) is:
\(OM = \frac{1}{2} \times 8 = 4\text{ cm}\)

The angle between the face \(VAB\) and the base \(ABCD\) is the angle \(\angle VMO\) in the right-angled triangle \(VOM\).
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{12}{4} = 3\)

\(\angle VMO = \arctan(3) \approx 71.565^\circ\)

To 1 decimal place, the angle is \(71.6^\circ\).

(a)
- M1: For calculating the diagonal of the base \(AC = \sqrt{128}\) or half-diagonal \(AO = \sqrt{32} \approx 5.66\).
- A1: For finding \(VA = \sqrt{176} \approx 13.3\text{ cm}\) (accept range 13.2 to 13.3).

(b)
- M1: For identifying the correct angle \(\angle VMO\) and finding the distance \(OM = 4\text{ cm}\).
- M1: For using trigonometry correctly, e.g., \(\tan(\theta) = \frac{12}{4}\).
- A1: For the final angle of \(71.6^\circ\) (accept range \(71.5^\circ\) to \(71.6^\circ\)).

(a) The sum of the first term \(S_1\) is equal to the first term \(a\):
\(a = S_1 = 2(1)^2 + 5(1) = 2 + 5 = 7\)

The sum of the first two terms is \(S_2\):
\(S_2 = 2(2)^2 + 5(2) = 8 + 10 = 18\)

The second term of the series, \(u_2\), is:
\(u_2 = S_2 - S_1 = 18 - 7 = 11\)

The common difference \(d\) is the difference between the second term and the first term:
\(d = u_2 - a = 11 - 7 = 4\)

(b) The formula for the \(n\)th term of an arithmetic series is \(u_n = a + (n - 1)d\).
For the 20th term:
\(u_{20} = a + 19d\)
\(u_{20} = 7 + 19(4) = 7 + 76 = 83\)

(a)
- M1: For finding the first term \(a = S_1 = 7\).
- M1: For finding \(S_2 = 18\) and subtracting \(S_1\) to find the second term \(11\).
- A1: For the correct common difference \(d = 4\).

(b)
- M1: For using the formula \(u_{20} = a + 19d\) or \(S_{20} - S_{19}\).
- A1: For the correct 20th term of \(83\).

To clear the fractions, multiply both sides of the equation by the common denominator \((x-3)(x+2)\):
\(4(x+2) + 3(x-3) = 1(x-3)(x+2)\)

Expand both sides:
\(4x + 8 + 3x - 9 = x^2 - x - 6\)
\(7x - 1 = x^2 - x - 6\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\(x^2 - x - 7x - 6 + 1 = 0\)
\(x^2 - 8x - 5 = 0\)

Since this quadratic does not factorise easily, we use the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
with \(a = 1\), \(b = -8\), and \(c = -5\):

\(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-5)}}{2(1)}
= \frac{8 \pm \sqrt{64 + 20}}{2}
= \frac{8 \pm \sqrt{84}}{2}\)

Calculating the two values of \(x\):
\(x_1 = \frac{8 + 9.16515}{2} \approx 8.5826 \approx 8.58\) (to 3 s.f.)

\(x_2 = \frac{8 - 9.16515}{2} \approx -0.5826 \approx -0.583\) (to 3 s.f.)

- M1: For a method to eliminate the fractions, such as multiplying all terms by \((x-3)(x+2)\).
- M1: For correctly expanding both sides to obtain \(7x - 1\) and \(x^2 - x - 6\).
- A1: For obtaining the correct quadratic equation in standard form: \(x^2 - 8x - 5 = 0\).
- M1: For applying the quadratic formula or completing the square correctly to their three-term quadratic.
- A1: For both correct solutions to 3 significant figures: \(x = 8.58\) and \(x = -0.583\).

(a) \(TA\) and \(TD\) are tangents to the circle, and \(OA\) and \(OD\) are radii.
The angle between a tangent and a radius is \(90^\circ\), so:
\(\angle OAT = 90^\circ\) and \(\angle ODT = 90^\circ\).

The angles in the quadrilateral \(OATD\) must add up to \(360^\circ\):
\(\angle AOD + \angle OAT + \angle ODT + \angle ATD = 360^\circ\)
\(\angle AOD + 90^\circ + 90^\circ + 54^\circ = 360^\circ\)
\(\angle AOD = 360^\circ - 234^\circ = 126^\circ\).

(b) Since \(B\) is on the major arc \(AD\), the angle \(\angle ABD\) is subtended at the circumference by the minor arc \(AD\). The angle subtended by the same arc at the centre is \(\angle AOD = 126^\circ\).

The angle subtended by an arc at the centre is twice the angle subtended at the circumference. Therefore:
\(\angle ABD = \frac{1}{2} \angle AOD = \frac{126^\circ}{2} = 63^\circ\).

- M1: For identifying that \(\angle OAT = \angle ODT = 90^\circ\) with the reason 'angle between tangent and radius is 90 degrees'.
- A1: For \(\angle AOD = 126^\circ\) with the reason 'sum of angles in a quadrilateral is 360 degrees'.
- M1: For dividing \(\angle AOD\) by 2 to find \(\angle ABD\).
- A1: For \(\angle ABD = 63^\circ\).
- C1: For providing both correct circle/geometry reasons ('angle between tangent and radius is 90 degrees' and 'angle at centre is twice angle at circumference') at relevant steps.

To find the velocity \(v\), we differentiate the displacement \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = 6t^2 - 30t + 24\)

To find when the velocity is equal to \(0\), we set \(v = 0\):

\(6t^2 - 30t + 24 = 0\)

Divide the entire equation by 6:

\(t^2 - 5t + 4 = 0\)

Factorise the quadratic equation:

\((t - 1)(t - 4) = 0\)

This gives \(t = 1\) and \(t = 4\).

The velocity is first equal to \(0\) at the instant \(t = 1\) second.

To find the acceleration \(a\), we differentiate the velocity \(v\) with respect to \(t\):

\(a = \frac{dv}{dt} = 12t - 30\)

Substitute \(t = 1\) into the expression for acceleration:

\(a = 12(1) - 30 = -18\text{ m/s}^2\).

M1: For differentiating the expression for displacement to find the velocity expression (at least two terms correct), e.g., \(v = 6t^2 - 30t + 24\).
M1: For setting their velocity expression equal to \(0\) and attempting to solve the quadratic equation.
A1: For identifying that the first instant the velocity is zero is at \(t = 1\).
M1: For differentiating their velocity expression to find the acceleration expression, e.g., \(a = 12t - 30\).
A1: For the final answer of \(-18\).

Let \(r\) be the initial number of red counters and \(g\) be the number of green counters in the bag.

Initially, the probability of choosing a red counter is:

\(\frac{r}{r+g} = \frac{3}{7}\)

Multiplying both sides by \(7(r+g)\) gives:

\(7r = 3(r+g)\)
\(7r = 3r + 3g\)
\(4r = 3g\) --- (Equation 1)

When 4 red counters are added, the new number of red counters is \(r + 4\) and the new total number of counters is \(r + g + 4\). The new probability is:

\(\frac{r+4}{r+g+4} = \frac{1}{2}\)

Multiplying both sides by \(2(r+g+4)\) gives:

\(2(r+4) = r + g + 4\)
\(2r + 8 = r + g + 4\)
\(r - g = -4\) --- (Equation 2)

From Equation 1, we can write \(r = \frac{3}{4}g\). Substituting this into Equation 2:

\(\frac{3}{4}g - g = -4\)
\(-\frac{1}{4}g = -4\)
\(g = 16\)

To find the number of red counters:

\(r = \frac{3}{4}(16) = 12\)

Thus, the number of green counters in the bag is 16.

M1: For setting up an equation for the initial probability, e.g., \(\frac{r}{r+g} = \frac{3}{7}\) or \(7r = 3(r+g)\).
M1: For setting up an equation for the new probability, e.g., \(\frac{r+4}{r+g+4} = \frac{1}{2}\) or \(2(r+4) = r+g+4\).
M1: For a valid method to solve the simultaneous equations to eliminate one variable, e.g., substituting \(r = 0.75g\) into the second equation.
A1: For finding either \(r = 12\) or \(g = 16\).
A1: For the final answer of 16.

First, use the volume of the sphere to find \(r\):

\(\frac{4}{3}\pi r^3 = 36\pi\)

Divide both sides by \(\pi\):

\(\frac{4}{3}r^3 = 36\)
\(r^3 = 36 \times \frac{3}{4} = 27\)
\(r = 3\text{ cm}\)

Next, write expressions for the surface areas:
- Surface area of the sphere: \(A_{\text{sphere}} = 4\pi r^2\)
- Total surface area of the cylinder (radius \(3r\), height \(h\)):
\(A_{\text{cylinder}} = 2\pi (3r)^2 + 2\pi (3r)h = 18\pi r^2 + 6\pi rh\)

We are given that \(A_{\text{cylinder}} = 6 \times A_{\text{sphere}}\):

\(18\pi r^2 + 6\pi rh = 6(4\pi r^2)\)
\(18\pi r^2 + 6\pi rh = 24\pi r^2\)

Subtract \(18\pi r^2\) from both sides:

\(6\pi rh = 6\pi r^2\)

Divide both sides by \(6\pi r\):

\(h = r\)

Since \(r = 3\), we have \(h = 3\text{ cm}\). The cylinder has a radius of \(3r = 9\text{ cm}\).

Now, calculate the volume of the cylinder:

\(V_{\text{cylinder}} =
\pi R^2 h = \pi (9)^2 (3) = 81 \times 3 \times \pi = 243\pi\text{ cm}^3\).

M1: For equating \(\frac{4}{3}\pi r^3 = 36\pi\) and finding \(r = 3\).
M1: For expressing the total surface area of the cylinder as \(2\pi(3r)^2 + 2\pi(3r)h\) or \(18\pi r^2 + 6\pi rh\).
M1: For writing a correct equation connecting the surface areas: \(18\pi r^2 + 6\pi rh = 6 \times (4\pi r^2)\), and simplifying to find \(h = r\) (or \(h = 3\)).
M1: For substituting their values into the formula for the volume of a cylinder, \(\pi (3r)^2 h\).
A1: For the final answer of \(243\pi\).

First, find an expression for \(g(3)\):

\(g(3) = 3(3) + k = 9 + k\)

Now, substitute this into \(f(x)\) to get \(fg(3)\):

\(fg(3) = f(9 + k) = \frac{6}{(9 + k) - 2} = \frac{6}{k+7}\)

We are given that \(fg(3) = 2\):

\(\frac{6}{k+7} = 2\)
\(6 = 2(k+7)\)
\(3 = k+7\)
\(k = -4\)

Now we need to find \(f^{-1}(k)\), which is \(f^{-1}(-4)\).
Let \(f^{-1}(-4) = y\), which means \(f(y) = -4\):

\(\frac{6}{y-2} = -4\)
\(6 = -4(y-2)\)
\(6 = -4y + 8\)
\(4y = 2\)
\(y = 0.5\)

Thus, \(f^{-1}(k) = 0.5\).

M1: For substituting \(x = 3\) into \(g(x)\) to get \(3(3)+k\) or \(9+k\).
M1: For writing a correct expression for \(fg(3)\), i.e., \(\frac{6}{(9+k)-2}\).
A1: For solving \(\frac{6}{k+7} = 2\) to get \(k = -4\).
M1: For a correct method to find \(f^{-1}(-4)\), e.g., setting \(\frac{6}{y-2} = -4\) and attempting to solve for \(y\), or finding \(f^{-1}(x) = \frac{6}{x} + 2\) and substituting \(x = -4\).
A1: For the final answer of \(0.5\) (or \(\frac{1}{2}\)).

Let \(O\) be the center of the square base \(ABCD\).
Let \(M\) be the midpoint of the side \(AB\).

The angle between the face \(VAB\) and the base \(ABCD\) is the angle \(\angle VMO\) in the right-angled triangle \(VOM\), where the vertical height \(VO\) is perpendicular to the base.

First, consider the face \(VAB\), which is an isosceles triangle with sides \(VA = VB = 13\text{ cm}\) and base \(AB = 10\text{ cm}\).
Since \(M\) is the midpoint of \(AB\), \(AM = 5\text{ cm}\).
In the right-angled triangle \(VAM\) (with angle \(\angle VMA = 90^\circ\)):

\(VM^2 + AM^2 = VA^2\)
\(VM^2 + 5^2 = 13^2\)
\(VM^2 + 25 = 169\)
\(VM^2 = 144\)
\(VM = 12\text{ cm}\)

Now consider the right-angled triangle \(VOM\) inside the pyramid.
Since \(O\) is the center of the square and \(M\) is the midpoint of \(AB\), the distance \(OM\) is half the side length of the square base:

\(OM = 10 \div 2 = 5\text{ cm}\)

In the right-angled triangle \(VOM\) (with angle \(\angle VOM = 90^\circ\)):

\(\cos(\angle VMO) = \frac{OM}{VM} = \frac{5}{12}\)

Now, calculate the angle \(\angle VMO\):

\(\angle VMO = \arccos\left(\frac{5}{12}\right) \approx 65.37568^\circ\)

Rounding to 1 decimal place gives \(65.4^\circ\).

M1: For identifying a correct right-angled triangle to find the slant height of face \(VAB\) (e.g., triangle \(VAM\), where \(M\) is the midpoint of \(AB\)).
M1: For a correct calculation to find this slant height \(VM = \sqrt{13^2 - 5^2} = 12\).
M1: For identifying that the horizontal distance from the midpoint of \(AB\) to the center of the base is \(5\text{ cm}\).
M1: For using a correct trigonometric ratio in triangle \(VOM\) to find the angle, e.g., \(\cos(\theta) = \frac{5}{12}\) or finding the vertical height \(VO = \sqrt{119}\) and using \(\tan(\theta) = \frac{\sqrt{119}}{5}\).
A1: For the final answer of \(65.4\) (accept values in the range 65.3 to 65.4).